A super elevation of a railroad track is the number of inches that the outside rail is raised to prevent side thrust on the wheel of flanges of cars rounding the curve at rated speed. Determine the super elevation e for track having a gauge of 4 ft and 8.5 inches of 2000 ft radius and a rated speed of 60 mph. What is the flange pressure P on the wheels of a 100,000 car that rounds the curve at 80 mph?
show solution step by step thanks
e=Gv^2/(gr)
e= (4.7)(88)/[(32.2)(2000)]
e= (0.565ft)(12in/1ft)
e=6.78in
Nagella
Answer
Oh, railroads and their super elevation! Let's chug along step by step to find the solution:
Step 1: Convert the gauge to inches.
The gauge of 4 ft and 8.5 inches can be calculated as 4 * 12 + 8.5 = 56.5 inches.
Step 2: Convert the rated speed to feet per second.
Since 1 mile has 5,280 feet and there are 60 minutes in an hour, the conversion is:
60 mph * (5280 ft / 1 mile) * (1 hour / 60 minutes) * (1 minute / 60 seconds) = 88 feet per second.
Step 3: Find the super elevation.
The formula for super elevation (e) is given by:
e = (v^2) / (g * r),
where v is the velocity in feet per second, g is the acceleration due to gravity (approximately 32.17 ft/s^2), and r is the radius of the curve in feet.
Plugging in the values:
e = (88^2) / (32.17 * 2000),
e ≈ 0.122 feet.
Step 4: Convert the super elevation to inches.
To get the super elevation in inches, we multiply by 12 inches per foot:
0.122 feet * 12 inches/foot ≈ 1.464 inches.
So, the super elevation is approximately 1.464 inches.
Now, let's take a spin to find the flange pressure!
Step 5: Convert the car's speed to feet per second.
Using the same conversion as before:
80 mph * (5280 ft / 1 mile) * (1 hour / 60 minutes) * (1 minute / 60 seconds) = 117.33 feet per second.
Step 6: Calculate the flange pressure.
The formula for flange pressure (P) is given by:
P = m * v^2 / r,
where m is the mass of the car and v is the velocity in feet per second, and r is the radius of the curve in feet.
Plugging in the values:
P = 100,000 lb * 117.33^2 ft^2 / 2000 ft,
P ≈ 686,315 lb.
Whew! So, the flange pressure on the wheels of the 100,000 lb car rounding the curve at 80 mph is approximately 686,315 lb.
And there you have it! Super elevation and flange pressure, all with a little rail-side humor.
To determine the super elevation (e) for a railroad track, we can use the formula:
e = (V^2 / (32.17 * R)) * 12
Where:
- e is the super elevation in inches
- V is the rated speed in mph
- R is the radius of the curve in feet
In this case, we have:
- Gauge = 4ft 8.5 inches = 4.71 feet
- Radius (R) = 2000 ft
- Rated speed (V) = 60 mph
Let's substitute these values into the formula to find the super elevation (e):
e = (60^2 / (32.17 * 2000)) * 12
e = (3600 / 64340) * 12
e ≈ 0.0663 * 12
e ≈ 0.7956 inches
Therefore, the super elevation (e) for this railroad track is approximately 0.7956 inches.
Next, to find the flange pressure (P) on the wheels of a car rounding the curve at 80 mph, we can use the formula:
P = (w * V^2) / (32.17 * R)
Where:
- P is the flange pressure in pounds per square inch (psi)
- w is the weight of the car in pounds
- V is the speed of the car in mph
- R is the radius of the curve in feet
Assuming the weight of the car (w) is 100,000 pounds and the speed (V) is 80 mph, let's substitute these values into the formula to find the flange pressure (P):
P = (100000 * 80^2) / (32.17 * 2000)
P = (100000 * 6400) / 64340
P = 640000000 / 64340
P ≈ 9952 psi
Therefore, the flange pressure (P) on the wheels of a 100,000-pound car rounding the curve at 80 mph is approximately 9952 psi.